Question 1131925
As {{{A}}},{{{B}}} and {{{C}}} are angles of a triangle, we have {{{A+B+C=180}}} 

and 

{{{A+B=180-C}}}

 or 

{{{(A+B)/2=90-C/2}}}


Hence 


{{{cos(A)+cos(B)+cos(C)}}}


={{{2cos(A+B)/2)*cos(A-B)/2) +1-2sin^2(C/2)-1}}}.......since {{{(A+B)/2=90-C/2}}}


={{{2cos(90-C/2)*cos((A-B)/2)+1 -2sin^2(C/2)}}}...since {{{cos(90-C/2)=sin(C/2)}}}


={{{2sin(C/2)*cos((A-B)/2)+1 -2sin^2(C/2)}}}......factor


={{{2sin(C/2)(cos((A-B)/2)-sin(C/2))+1}}} ........from {{{sin((A+B)/2)=sine(90-C/2)}}}=>{{{sine((A+B)/2)=sin(C/2)}}}


={{{2sin(C/2)(cos((A-B)/2)-sin((A+B)/2))+1}}}


={{{2sin(C/2)(2sin(A/2)*sin(B/2))+1}}}


={{{1+4sin(A/2)*sin(B/2)sin(C/2)}}}